Answer
$g'(x)= \frac{1-3x}{e^{3x}}$
Work Step by Step
$g(x)=\frac{x}{e^{3x}} = (x)(e^{-3x})$
Using Product Rule and $\frac{d}{dx}e^{nx} = ne^{nx}$
$g'(x)=(1)(e^{-3x})+(x)(-3e^{-3x}) = \frac{1}{e^{3x}} - \frac{3x}{e^{3x}} = \frac{1-3x}{e^{3x}}$
Calculus: Early Transcendentals (2nd Edition)
by Briggs, Bill L.; Cochran, Lyle; Gillett, Bernard
Published by
Pearson
ISBN 10:
0321947347
ISBN 13:
978-0-32194-734-5
Chapter 3 - Derivatives - 3.4 The Product and Quotient Rules - 3.4 Exercises - Page 160: 47
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